Answer:
[tex]7 < k < 13[/tex]
[tex]-6>y>-12[/tex]
[tex]m > 5[/tex]
[tex]3 < p < \frac{15}{2}[/tex]
Step-by-step explanation:
Given
[tex]5 < k -2 < 11[/tex]
[tex]-4 > y + 2 > -10[/tex]
[tex]5 -m < 4\ or\ 7m > 35[/tex]
[tex]3 < 2p -3 < 12[/tex]
Required
Solve
Solving (1):
[tex]5 < k -2 < 11[/tex]
The above can be represented as:
[tex]5<k -2[/tex] and [tex]k -2<11[/tex]
Solving [tex]5<k -2[/tex]
[tex]5 + 2 <k[/tex]
[tex]7<k[/tex]
Solving [tex]k -2<11[/tex]
[tex]k < 2 + 11[/tex]
[tex]k < 13[/tex]
So, we have:
[tex]7<k[/tex] and [tex]k < 13[/tex]
This can be combined as:
[tex]7 < k < 13[/tex]
Solving (2):
[tex]-4 > y + 2 > -10[/tex]
Split as:
[tex]-4 > y + 2[/tex] and [tex]y + 2 > -10[/tex]
Solving [tex]-4 > y + 2[/tex]
[tex]-4 - 2 > y[/tex]
[tex]-6 > y[/tex]
Solving [tex]y + 2 > -10[/tex]
[tex]y > -10 - 2[/tex]
[tex]y > -12[/tex]
So, we have:
[tex]-6 > y[/tex] and [tex]y > -12[/tex]
This gives
[tex]-6>y>-12[/tex]
Solving (3):
[tex]5 -m < 4\ or\ 7m > 35[/tex]
Solve for m in both cases:
[tex]5 - m < 4[/tex]
[tex]-m<4 - 5[/tex]
[tex]-m<- 1[/tex]
[tex]m > 1[/tex]
[tex]7m > 35[/tex]
Divide both sides by 5
[tex]m > 5[/tex]
So, we have:
[tex]m > 1[/tex] or [tex]m > 5[/tex]
The solution is [tex]m > 5[/tex] because [tex]5> 1[/tex]
Solving (4):
[tex]3 < 2p -3 < 12[/tex]
The above can be represented as:
[tex]3 < 2p - 3[/tex] and [tex]2p - 3<12[/tex]
Solving [tex]3 < 2p - 3[/tex]
Add 3 to both sides
[tex]3 + 3 < 2p - 3 + 3[/tex]
[tex]6 < 2p[/tex]
Divide both sids by 2
[tex]3 < p[/tex]
[tex]2p - 3<12[/tex]
Add 3 to both sides
[tex]2p - 3 + 3 < 12 + 3[/tex]
[tex]2p < 12 + 3[/tex]
[tex]2p < 15[/tex]
Solve for p
[tex]p < \frac{15}{2}[/tex]
So, we have:
[tex]3 < p[/tex] and [tex]p < \frac{15}{2}[/tex]
Combined as:
[tex]3 < p < \frac{15}{2}[/tex]
